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Topic: constructing an example
Replies: 6   Last Post: Jul 28, 2011 11:39 AM

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Rob Johnson

Posts: 1,771
Registered: 12/6/04
Re: constructing an example
Posted: Jul 27, 2011 4:57 PM
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In article <folt27pu1l3bv9efboq189203ucr37pl79@4ax.com>,
David C. Ullrich <ullrich@math.okstate.edu> wrote:
>On Mon, 25 Jul 2011 11:22:35 -0700 (PDT), Dan <dan.ms.chaos@gmail.com>
>wrote:
>

>>Let g : [0,1) -> R be
>>g(x) = -1/2 for x <= 1/2
>>g(x) = 1/2 for x> 1/2
>>
>>u(x) = (2*x ) mod 1
>>
>>Can h(x) be constructed so that h(u(x)) - h(x) is equivalent to g(x)
>>in LP2 ?
>>(integral from 0 to 1 over x of ((h(u(x)) - h(x) ) - g(x) )^2 =
>>0 )

>
>One can save typing by replacing your g by 2g.
>
>So the question is does there exist h such that h(2x) - h(x) = -1
>almost everywhere for 0 < x < 1/2 and h(2x - 1) - h(x) = 1
>almost everywhere for 1/2 < x < 1.
>
>It seems to me that there is no such h in L^2 (I gather your
>"LP2" is the space commonly known as L^2?). If you say
>c_n is the n-th Fourier coefficient of h then it seems to me
>that (dropping irrelevant constants) you'd need
>c_n = 1/n for n odd and c_{2n} = c_n.
>
>Except that c_0 is arbitrary, there's exactly one sequence
>that satisfies those two conditions. It doesn't tend to 0,
>so h cannot be L^2 or even L^1. On the other hand,
>that sequence is bounded, so there _is_ a 2pi-periodic
>diastribution that does what you ask, at least formally.
>Whether that distribution can be given as a locally
>integrable function, or whether h can be taken to be
>measurable and satisfy your condition pointwise, is
>not immediately clear to me.


If we don't drop the irrelevant constants, we get that the
fourier series for h(x) would formally be

+oo
2 ---
h(x) = - > c sin(2?kx)
? --- k
k=1

where c_k is the reciprocal of the largest odd divisor of k.

Since the fourier series of g(x) is

+oo
2 --- 1
g(x) = - - > ---- sin(2?(2k+1)x)
? --- 2k+1
k=0

We get that, formally,

+oo
--- k
h(x) = - > g(2 x)
---
k=0

Here, at least formally, h(2x) - h(x) = g(x).

If we try to compute h(x) for a given x which doesn't have a
finite binary expansion, we get that h(x) is the sum over the
binary digits of x, 1/2 for 0 and -1/2 for 1. My bet is against
there being a locally integrable function for h.

Consider h as the limit as n -> oo of

n
--- k
h (x) = - > g(2 x)
n ---
k=0

Then, it is pretty straightforward to show that h is bounded as a
distribution on Lip(?) for ? > 0, with a constant of 1/(2^? - 1),
which is about 1/(log(2) ?) when ? is near 0.

Rob Johnson <rob@trash.whim.org>
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